Distances in an expanding universe

Builds on: Sections 13.2, 13.4 and 4.3
Reading time: 45 min
Exercises: 3
Olympiad problems: 4

The Universe is about 13.5 billion years old, yet the furthest matter whose light could have reached us is now about 45 billion light-years away. Nothing has moved faster than light through space. The space between has stretched while the light travelled, and the source has gone on moving away. In an expanding universe the word "distance" can mean several different things, and each answers a different question. In this section we meet the three that olympiad problems use and see how to find each one from a redshift.

13.6.1The distance now

Before you readMake a guess first

A galaxy is far enough away that the Hubble–Lemaître law gives it a recession speed of 1.5 times the speed of light. Can we see it?

The distance to a galaxy today, measured along a line through space at this moment, as if with a chain of rulers laid end to end, is its proper distance now. It is also called the comoving distance, because for a galaxy that only moves with the expansion it is fixed when distances are measured in today's units. Light reaching us from redshift $z$ crossed this distance while the Universe grew, and each stretch of its path counts at the rate of expansion at the time it was crossed.

ResultThe comoving distance(13.12)

D_{C} = c \int_{0}^{z} \frac{d z^{'}}{H (z^{'})} = \frac{c}{H_{0}} \int_{0}^{z} \frac{d z^{'}}{E (z^{'})}

$D_{C}$	the comoving distance to a source at redshift $z$ , its proper distance today
$H (z)$	the Hubble parameter at redshift $z$
$E (z)$	the ratio $H / H_{0}$ from (13.8), $\sqrt{Ω_{r} (1 + z)^{4} + Ω_{m} (1 + z)^{3} + Ω_{k} (1 + z)^{2} + Ω_{Λ}}$

For small

z

E \approx 1

and

D_{C} \approx c z / H_{0}

, the simple law of Section 13.1. The present recession speed of the source is

v = H_{0} D_{C}

, which passes the speed of light beyond the Hubble distance

c / H_{0}

, at

z \approx 1.5

in our Universe. Nothing moves through space faster than light, since this speed is the growth of the space between.

For a source at $z = 1$ in our model the integral gives $D_{C} = 3304$ Mpc, against 4283 Mpc from $c z / H_{0}$ . The source now recedes at 0.77 times the speed of light. Taken out to the Big Bang, the same integral gives the distance of the furthest matter we could ever have seen, 13.9 Gpc, or about 45 billion light-years.

13.6.2Brightness and size

A telescope measures fluxes and angles, not distances. In ordinary space, flux falls as $1 / d^{2}$ ((4.1)) and an object of size $ℓ$ covers an angle $ℓ / d$ . To keep these familiar rules, cosmologists define two more distances, chosen so that the rules stay true.

Light from a source at redshift $z$ spreads over a sphere of radius $D_{C}$ by the time it arrives. Each photon has also lost energy by $1 + z$ , and the photons arrive more slowly by another factor of $1 + z$ ((13.4)), so the flux is less than $L / 4 π D_{C}^{2}$ by $(1 + z)^{2}$ . The light set out when all lengths were smaller by $1 + z$ , so an object of size $ℓ$ then covered a larger angle than its present comoving distance would suggest.

ResultThe luminosity and angular-diameter distances(13.13)

\begin{aligned} F & = \frac{L}{4 π D_{L}^{2}}, D_{L} = (1 + z) D_{C} \\ θ & = \frac{ℓ}{D_{A}}, D_{A} = \frac{D_{C}}{1 + z} \end{aligned}

$D_{L}$	the luminosity distance, which gives the flux $F$ from the luminosity $L$
$D_{A}$	the angular-diameter distance, which gives the angle $θ$ covered by an object of size $ℓ$
$D_{C}$	the comoving distance, for flat space

The two are linked by

D_{L} = (1 + z)^{2} D_{A}

. The distance modulus of (4.4) uses

D_{L}

. In curved space

D_{C}

in both formulas is replaced by a slightly different distance, which this course does not need.

Figure 13.5Distances to a redshift

Matter

Ω_{m}

Dark energy

Ω_{Λ}

Redshift

z

Space is
Comoving distance
Luminosity distance
Angular-diameter distance
Light left

The figure shows the comoving distance (heavy), the luminosity distance (thin) and the angular-diameter distance (dashed) against redshift for a model universe, with the simple law

c z / H_{0}

dotted. All four agree at small redshifts. The luminosity distance soon runs far ahead, while the angular-diameter distance reaches a largest value near

z = 1.6

and then falls, so beyond that redshift a galaxy of a given size looks larger the further away it is. The table gives the distances at the chosen redshift and when the light set out. The figure can be set to , , or , in which every distance is smaller.

Quick checkThe distance modulus at redshift 1

A source at $z = 1$ has a comoving distance of 3304 Mpc. What is its distance modulus?

Example 13.9How large a galaxy at redshift 1 looks

A galaxy 10 kpc across is seen at $z = 1$ , where $D_{C} = 3304$ Mpc. What angle does it cover?

Solution

By (13.13), $D_{A} = 3304 / 2 = 1652$ Mpc, so $θ = 0.010 / 1652 = 6.05 \times 10^{- 6}$ rad.

AnswerThe galaxy covers 1.25″, about the size of a star blurred by the atmosphere, which is why the Hubble and Webb telescopes, above the air, are needed to see the shapes of distant galaxies.

Common errorsSection 13.6

The law $d = c z / H_{0}$ holds only for small redshifts. At $z = 1$ it is already about 30 per cent too large in our Universe.
The flux of a distant source uses the luminosity distance, and its apparent size the angular-diameter distance. Using one in place of the other is wrong by $(1 + z)^{2}$ .
A recession speed faster than light does not break relativity. Relativity limits motion through space, and the recession is the growth of space between the galaxies.
The redshift cannot be turned into a distance without a model of the Universe. Problems that give no model often mean the simple law for small $z$ , or a stated model, and the model used should be said.

ExercisesSection 13.63 questions

13.6.1
In our model the comoving distance to $z = 2$ is about 5180 Mpc. What is the luminosity distance?
Mpc
13.6.2
A galaxy is seen at $z = 3$ . What is its angular-diameter distance as a fraction of its comoving distance?
13.6.3
Why is the flux from a distant galaxy smaller than $L / 4 π D_{C}^{2}$ ?

Olympiad problemsPast IOAA problems, restated4 problems

2018Faster than light10 points

Adapted from IOAA 2018 (Beijing), theory problem 1. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.

Say whether each statement is true or false.

(a) For some galaxies the apparent recession speed is greater than the speed of light. [2 points]

(b) The relation between speed and distance given by Hubble does not allow recession speeds greater than the speed of light. [2 points]

(d) If some galaxies had apparent recession speeds greater than the speed of light, their photons could never reach us. [2 points]

(e) Because the expansion is speeding up, photons sent out now by galaxies whose apparent recession speed equals the speed of light will never reach us. [2 points]

I solved it without the solution

AnswerTrue, false, true, false, false.

2013Three ways to turn a redshift into a speed10 points

Adapted from IOAA 2013 (Volos), theory problem 15. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.

The redshifts of three distant radio sources are 0.536 for 3C 279, 1.029 for 3C 245 and 3.8 for 4C 41.17.

(a) Find the apparent recession speed of each, using (1) the classical rule $v = c z$ , (2) the approximate cosmological rule $v = c \ln (1 + z)$ , and (3) the relativistic Doppler formula. [5 points]

(b) For each rule, give the speeds as percentages of the speed of light. [3 points]

I solved it without the solution

Answer(a, b) Classical 54%, 103% and 380%. Logarithmic 43%, 71% and 157%. Relativistic 41%, 61% and 92%. (c) The approximate cosmological rule is the most suitable of the three, and the true speeds in our Universe, 47%, 79% and 164%, lie closest to it.

2022A journey between galaxies10 points

Adapted from IOAA 2022 (Kutaisi), theory problem 4. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.

We set out for a planet that is $d_{0} = 10$ Mpc away when we leave. The Universe keeps expanding by the Hubble–Lemaître law, and the Hubble constant $H = 70$ km/s/Mpc does not change.

(a) Write the distance between Earth and the planet at a time $t$ after we leave. [2 points]

(b) What is the smallest constant speed $v_{0}$ of the rocket, relative to the space it passes through, that will reach the planet? Can a rocket at $v_{0} = 1000$ km/s reach it? If so, how long does the journey take? [8 points]

I solved it without the solution

Answer(a) $d = d_{0} e^{H t}$ . (b) The rocket needs more than 700 km/s, so yes, at 1000 km/s it arrives after about 16.8 billion years. The official solution misprints the smallest speed as 70 km/s and the time as $5.3 \times 10^{16}$ s, though its 17 billion years is right.

2010The distance to a quasar10 points

Adapted from IOAA 2010 (Beijing), theory problem 3. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.

A quasar has a redshift of $z = 0.20$ . Estimate its distance, taking the Hubble constant as 72 km/s/Mpc.

I solved it without the solution

AnswerThe quasar is about 750 to 800 Mpc away. The official solution accepts 750 Mpc from the Doppler route or about 735 Mpc from a flat universe of matter alone. Our model gives 794 Mpc. For a flat universe of matter alone the formula gives 726 Mpc, a little less than the official 735 Mpc.